The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad A(t) = 415 - \sin\left(\dfrac{2\pi (t + 23.2)}{92.8}\right)A(t)=415−sin( 92.8 2π(t+23.2) )space, A, left parenthesis, t, right parenthesis, equals, 415, minus, sine, left parenthesis, start fraction, 2, pi, left parenthesis, t, plus, 23, point, 2, right parenthesis, divided by, 92, point, 8, end fraction, right parenthesis. The International Space Station reaches its perigee once in every orbit. How long does the International Space Station take to orbit the earth? Give an exact answer.

Question

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Respuesta :

shahnoorazhar3 shahnoorazhar3 · Answer 1 · 2020-02-07T01:07:52+01:00

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

[tex]A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex]

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

[tex]A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )[/tex]

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

T = 2*p / w

T = 2*p / (2*p / 92.8)

Hence, T = 92.8 min

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-19T12:49:02+02:00

The time of the International Space Station is mathematically given as

T = 5568sec

Generally, the equation for the altitude of the International Space Station is mathematically given as

[tex]A(t) \left = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex]

Therefore

Using the equation

[tex]A(t) \left = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex] we decipher time to be

T = 2*p / w

In conclusion

T = 2*p / (2*p / 92.8)

T = 5568sec